1 Answer
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Without more information, this is only one possible solution. Let us suppose the curve is modelled by an exponential
$$f(x) = a\exp(bx)$$
Then we have
$$f(2) = a\exp(2b) = 2600,\ \ \ f(7)=a\exp(7b)=5230$$
Solving for the coefficients gives us
$$\exp(5b) = \frac{5230}{2600} \implies b = \frac{1}{5}\ln\left(\frac{523}{260}\right)\approx 0.1398$$
Similarly, we find that $a$ is given by
$$a = \frac{2600}{\exp(2b)}=1965.9$$
This gives
$$f(x) = 1965.9\exp(0.1398x)$$